Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $a = \dfrac{2r - 4}{r^2 + 2r - 8} \div \dfrac{2r^2 + 18r}{-2r^3 - 14r^2 + 36r} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{2r - 4}{r^2 + 2r - 8} \times \dfrac{-2r^3 - 14r^2 + 36r}{2r^2 + 18r} $ First factor out any common factors. $a = \dfrac{2(r - 2)}{r^2 + 2r - 8} \times \dfrac{-2r(r^2 + 7r - 18)}{2r(r + 9)} $ Then factor the quadratic expressions. $a = \dfrac {2(r - 2)} {(r - 2)(r + 4)} \times \dfrac {-2r(r - 2)(r + 9)} {2r(r + 9)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {2(r - 2) \times -2r(r - 2)(r + 9) } { (r - 2)(r + 4) \times 2r(r + 9)} $ $a = \dfrac {-4r(r - 2)(r + 9)(r - 2)} {2r(r - 2)(r + 4)(r + 9)} $ Notice that $(r - 2)$ and $(r + 9)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-4r\cancel{(r - 2)}(r + 9)(r - 2)} {2r\cancel{(r - 2)}(r + 4)(r + 9)} $ We are dividing by $r - 2$ , so $r - 2 \neq 0$ Therefore, $r \neq 2$ $a = \dfrac {-4r\cancel{(r - 2)}\cancel{(r + 9)}(r - 2)} {2r\cancel{(r - 2)}(r + 4)\cancel{(r + 9)}} $ We are dividing by $r + 9$ , so $r + 9 \neq 0$ Therefore, $r \neq -9$ $a = \dfrac {-4r(r - 2)} {2r(r + 4)} $ $ a = \dfrac{-2(r - 2)}{r + 4}; r \neq 2; r \neq -9 $